The sum S = sin theta + sin 2theta + dots + s | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The sum of the series is given by $S = \sum_{k=1}^{n} \sin(k	heta)$.Using the formula for the sum of sines in arithmetic progression: $\sum_{k=0}

Vedclass · Accepted Answer

$\frac{\sin \frac{1}{2}(n + 1)	heta \sin \frac{1}{2}n	heta}{\sin \frac{	heta}{2}}$

Vedclass · Answer

(A) The sum of the series is given by $S = \sum_{k=1}^{n} \sin(k	heta)$.Using the formula for the sum of sines in arithmetic progression: $\sum_{k=0}^{n-1} \sin(\alpha + k\beta) = \frac{\sin(n\beta/2)}{\sin(\beta/2)} \sin(\alpha + \frac{(n-1)\beta}{2})$.Here,$\alpha = 	heta$ and $\beta = 	heta$.Substituting these values,we get:$S = \frac{\sin(n	heta/2)}{\sin(	heta/2)} \sin(	heta + \frac{(n-1)	heta}{2})$$S = \frac{\sin(n	heta/2)}{\sin(	heta/2)} \sin(\frac{2	heta + n	heta - 	heta}{2})$$S = \frac{\sin(n	heta/2) \sin(\frac{(n+1)	heta}{2})}{\sin(	heta/2)}$.

The sum $S = \sin \theta + \sin 2\theta + \dots + \sin n\theta$ equals

Similar Questions

If $0 \leq A, B \leq \frac{\pi}{4}$ and $\cot A + \cot B + \tan A + \tan B = \cot A \cot B - \tan A \tan B$,then $\sin(A + B) = $

If $\cos \theta = -\frac{\sqrt{3}}{2}$ and $\sin \alpha = -\frac{3}{5}$,where $\theta$ does not lie in the third quadrant,then $\frac{2 \tan \alpha + \sqrt{3} \tan \theta}{\cot^2 \theta + \cos \alpha}$ is equal to

If $\frac{5 \sinh 2x}{7+6 \cosh 2x} = \frac{3}{2}$,then $3 \tanh^2 x + 20 \tanh x = $

The value of $\cos(18^{\circ}-A) \cdot \cos(18^{\circ}+A) - \cos(72^{\circ}-A) \cdot \cos(72^{\circ}+A)$ is:

If $\frac{\cos x}{a} = \frac{\cos (x + \theta)}{b} = \frac{\cos (x + 2\theta)}{c} = \frac{\cos (x + 3\theta)}{d}$,then $\left( \frac{a + c}{b + d} \right)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module